# Examples where roots are necessary for the solution

I currently write an article where I want to introduce roots. Thus I need to motivate them. Here I said, they can be used to find solutions of equations like $x^n=a$. Now I want to make some examples, where this problem arises. So far I found:

• What is the side length of a square with the area of $10 m^2$?
• What are the roots of the polynomial $x^2+x-1$?

What examples would you take? I do not have the feeling that mine are really convincing to study roots...

• What is the hypotenuse of the isosceles right triangle with leg length one inch? (Lots of history with this.) Commented May 28, 2015 at 21:43
• Are you including e.g. all 0s of polynomials? If so, what about Calculus problems where one wants the derivative (or second derivative, etc) of a polynomial (which is also a polynomial...) to be 0? I do not think there is a shortage of such problems. Maybe you could make your question a bit more specific? Commented May 28, 2015 at 22:26
• How about solving for the percentage annual return which will double an investment in ten years? Commented Jun 5, 2015 at 4:54

Ray-tracing, which underlies much of high-end computer graphics (from Toy Story to Frozen) relies on computing the intersection of a ray line-of-sight with a geometric object. For example, where does the ray $a + t v$ with parameter $t$, $a=(2,0)$, $v=(-1,\frac{1}{2})$, intersect the unit circle $x^2 + y^2 = 1$? Substitution of $a + t v = \left( 2-t,\frac{t}{2} \right)$ for $(x,y)$ in the circle equation leads to \begin{eqnarray} (2-t)^2 + \frac{t^2}{4} &=& 1 \;, \\ \frac{5 t^2}{4}-4 t+3 &=& 0 \end{eqnarray} whose roots are $t=2$, or $t=\frac{6}{5}$, yielding the intersection points \begin{eqnarray} (x,y) &=& (0,1) \;, \\ (x,y) &=& \left( \frac{4}{5},\frac{3}{5} \right) &=& (0.8, 0.6) \;. \end{eqnarray}

Because the latter point is closer to the ray origin $a$, that $(0.8, 0.6)$ point is the one whose surface characteristics determine the color/shade of the image pixel represented by the ray line-of-sight.

To compute the golden ratio, we need to find the positive root of a quadratic equation.

Two quantities $a$ and $b$ with $a>b>0$ are said to be in the golden ratio if $a$ is to $b$ as $a+b$ is to $a$.

(Image source: Wikipedia)

Algebraically, $$\frac a b= \frac {a+b} a.$$ If $\varphi=\tfrac a b$ denotes the golden ratio, then $$\varphi = 1+\varphi^{-1}.$$ Multiplying with $\varphi$ and rearranging gives the equation $$\varphi^2-\varphi-1=0,$$ which has a positive root $$\varphi=\frac {1+\sqrt 5} 2 \approx 1.618$$

• Let $x = -\varphi$. Then $\varphi^2 - \varphi - 1 = x^2 + x - 1$, which amounts to the second bullet-point from the OP. Commented Jun 7, 2015 at 5:12
• @BenjaminDickman ... or let $x=b/a=\varphi^{-1}$. Then you also end up with the equation $x^2+x-1=0$. Commented Jun 7, 2015 at 14:58